Point is chosen inside the square, find angle.

Question

Inside the square $MNPK$ point $O$ in such a way that $MO:ON:OP =1:2:3$. Find $\angle MON$. I tried to solve this problem with vectors, but I couldn't find coordinates of point $O$. Can someone explain this to me?

Answer 1

Let $A,B,C$ be the reflections of $O$ with respect to $PN,PM,MN$.
We have $\widehat{BMC}=\widehat{BPA}=90^\circ$ and $\widehat{CNA}=180^\circ$.
The sides of $ABC$ follow the proportion $$AB:AC:BC=\sqrt{2}\,OP:2\,ON:\sqrt{2}\,OM=3\sqrt{2}:4:\sqrt{2}$$ hence by the inverse Pythagorean theorem $\widehat{BCA}=90^\circ$ and $$\widehat{MON}=\widehat{MCA}=\widehat{MCB}+\widehat{BCA}=45^\circ+90^\circ=\color{red}{135^\circ}.$$

Answer 2

Station the square so that $M$ is the origin and $N$ lies along the positive $x$-axis. Assume the side of the square is $a$, which we will rescale so that $MO = 1$ and then4 $ON = 2$ and $OP = 3$. Let $O$ be located at $(x,y)$ then length of $MO$ implies $$ x^2 + y^2 = 1^2 $$ and length of $ON$ implies $$ (a-x)^2 + y^2 = 2^2. $$ Write down the implication of length of $OP$ and solve 3 equations in 3 unknowns to find $a,x,y$. You can then find the angle, for example, from the law of cosines.

Answer 3

Apply the Pythagorean theorem to establish the system of three equations

\begin{align} x^2+y^2=4,\>\>\>\>\> x^2+(a-y)^2=1,\>\>\>\>\> (a-x)^2+y^2=9 \end{align}

The 2nd and 3rd equations lead to $y=\frac{a^2+3}{2a}$ and $x=\frac{a^2-5}{2a}$, respectively. Plug them into the 1st equation to get $a^4 - 10a^2 +17=0$, which yields $a^2 =5+2\sqrt2$. Then, use the cosine rule for the triangle OMN to obtain

$$\cos\angle MON = \frac{OM^2+ON^2-a^2}{2|OM||ON|} = \frac{1+4-(5+2\sqrt2)}{2\cdot 1\cdot2}=-\frac{\sqrt2}2$$

Thus, $\angle MON =135^\circ$.